# How many atoms of oxygen are in one MOLE of Al_2(CrO_4)_3 8H_2O?

Well, there are clearly (?) 20 oxygen atoms per formula unit of $A {l}_{2} {\left(C r {O}_{4}\right)}_{3} \cdot 8 {H}_{2} O$. Agreed?
So thus in 1 mol of the aluminum chromate hydrate there must be 20 mol of oxygen atoms, i.e. $20 \times {N}_{A}$ $\text{oxygen atoms}$, where ${N}_{A} = \text{Avogadro's number}$ $=$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$. That's a lot of arithmetic.